Download 12:2: Applications of Maxima and Minima

12
Further Applications
of the Derivative
Copyright © Cengage Learning. All rights reserved.
12.2
Applications of Maxima and Minima
Copyright © Cengage Learning. All rights reserved.
Example 1 – Minimizing Average Cost
Gymnast Clothing manufactures expensive hockey jerseys
for sale to college bookstores in runs of up to 500. Its cost
(in dollars) for a run of x hockey jerseys is
C(x) = 2,000 + 10x + 0.2x2.
How many jerseys should Gymnast produce per run in
order to minimize average cost?
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Example 1 – Solution
Here is the procedure we will follow to solve problems like
this.
1. Identify the unknown(s). There is one unknown: x, the
number of hockey jerseys Gymnast should produce per
run.
2. Identify the objective function. The objective function
is the quantity that must be made as small (in this case)
as possible.
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Example 1 – Solution
cont’d
In this example it is the average cost, which is given by
3. Identify the constraints (if any). At most 500 jerseys
can be manufactured in a run. Also,
is not defined.
Thus, x is constrained by
0 < x ≤ 500.
Put another way, the domain of the objective function
is (0, 500].
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Example 1 – Solution
cont’d
4. State and solve the resulting optimization problem.
Our optimization problem is:
Objective function
subject to 0 < x ≤ 500.
Constraint
We first calculate
We solve
to find x = ±100. We reject x = –100
because –100 is not in the domain of (and makes no
sense), so we have one stationary point, at x = 100.
There, the average cost is
per jersey.
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Example 1 – Solution
cont’d
The only point at which the formula for is not defined is
x = 0, but that is not in the domain of so we have no
singular points. We have one endpoint in the domain, at
x = 500. There, the average cost is
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